Answer
The solution of the given question is $\ \sin \ x$.
Work Step by Step
$\frac{{{\sec }^{2}}\ x\ \csc \ x}{{{\sec }^{2\ }}x\ +\ {{\csc }^{2}}x}$
By using the reciprocal identity of trigonometry, which is $\text{sec }x=\frac{1}{\text{cos }x}$ and $\csc x=\ \frac{1}{\sin x}$ the expression can be further simplified as:
$\frac{{{\sec }^{2}}\ x\ \csc \ x}{{{\sec }^{2\ }}x\ +\ {{\csc }^{2}}x}=\ \frac{\frac{1}{{{\cos }^{2}}x}\times \frac{1}{\sin x}}{\frac{1}{{{\cos }^{2}}x}+\frac{1}{{{\sin }^{2}}x}}$
Now, the equation above can be further simplified by multiplying $\frac{{{\cos }^{2}}x{{\sin }^{2}}x}{{{\cos }^{2}}x{{\sin }^{2}}x}$
$\begin{align}
& \frac{{{\sec }^{2}}\ x\ \csc \ x}{{{\sec }^{2\ }}x\ +\ {{\csc }^{2}}x}=\ \frac{\frac{1}{{{\cos }^{2}}x}\times \frac{1}{\sin x}}{\frac{1}{{{\cos }^{2}}x}+\frac{1}{{{\sin }^{2}}x}}\times \frac{{{\cos }^{2}}x{{\sin }^{2}}x}{{{\cos }^{2}}x{{\sin }^{2}}x} \\
& =\frac{\sin }{{{\sin }^{2}}x+{{\cos }^{2}}x} \\
& =\frac{\sin \ x}{1} \\
& =\sin x
\end{align}$
Conjecture: Left side is equal to $\sin x$.
Thus, the left side of the expression is equal to the right side, which is $\frac{{{\sec }^{2}}\ x\ \csc \ x}{{{\sec }^{2\ }}x\ +\ {{\csc }^{2}}x}=\ \sin \ x$.
